Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{9q + 18}{q + 4} \div \dfrac{q^2 + 6q + 8}{-3q - 12} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{9q + 18}{q + 4} \times \dfrac{-3q - 12}{q^2 + 6q + 8} $ First factor the quadratic. $a = \dfrac{9q + 18}{q + 4} \times \dfrac{-3q - 12}{(q + 2)(q + 4)} $ Then factor out any other terms. $a = \dfrac{9(q + 2)}{q + 4} \times \dfrac{-3(q + 4)}{(q + 2)(q + 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ 9(q + 2) \times -3(q + 4) } { (q + 4) \times (q + 2)(q + 4) } $ $a = \dfrac{ -27(q + 2)(q + 4)}{ (q + 4)(q + 2)(q + 4)} $ Notice that $(q + 4)$ and $(q + 2)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -27\cancel{(q + 2)}(q + 4)}{ (q + 4)\cancel{(q + 2)}(q + 4)} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $a = \dfrac{ -27\cancel{(q + 2)}\cancel{(q + 4)}}{ \cancel{(q + 4)}\cancel{(q + 2)}(q + 4)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $a = \dfrac{-27}{q + 4} ; \space q \neq -2 ; \space q \neq -4 $